Answered

Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

In the oxidation of ethane: 2 C2H6 + 7 02 → 4 CO2 + 6 H2O how many
liters of O2 are required to react with 90 grams of ethane?

Sagot :

Eduard22sly

Answer:

10.5 L of O₂.

Explanation:

We'll begin by calculating the number of mole in 90 g of C₂H₆. This can be obtained as follow:

Molar mass of C₂H₆ = (12×2) + (6×1)

= 24 + 6

= 30 g/mol

Mass of C₂H₆ = 90 g

Mole of C₂H₆ =?

Mole = mass / molar mass

Mole of C₂H₆ = 90 / 30

Mole of C₂H₆ = 3 moles.

Finally, we shall determine the volume of O₂ required. This is illustrated:

2C₂H₆ + 7O₂ —> 4CO₂ + 6H₂O

From the balanced equation above, we can say that:

2 L of C₂H₆ required 7 L of O₂.

Therefore, 3 L of C₂H₆ will require

= (3 × 7)/ 2 = 10.5 L of O₂.

Thus, 10.5 L of O₂ is required for the reaction.

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.